Retarded Potentials and Jefimenko's Equations

Derivation

Retarded Potentials and Jefimenko's Equations

D-168 Home PU-204 Threads fields · waves Depends on Potentials and Gauge Freedom, Electromagnetic Wave Equation in Vacuum

Statement

Starting from the inhomogeneous wave equations that Maxwell's equations impose on the potentials in Lorenz gauge, \( \Box \varphi = -\rho/\varepsilon_0 \) and \( \Box \mathbf{A} = -\mu_0 \mathbf{J} \) with \( \Box \equiv \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \), we construct the retarded Green's function of the d'Alembertian and obtain the retarded potentials — each source element contributing at the retarded time \( t_r = t - |\mathbf{r}-\mathbf{r}'|/c \). Differentiating these potentials, \( \mathbf{E} = -\nabla\varphi - \partial_t \mathbf{A} \) and \( \mathbf{B} = \nabla\times\mathbf{A} \), yields Jefimenko's equations: the exact, manifestly causal fields of an arbitrary prescribed charge and current distribution in vacuum.

Why it matters

This is the general solution of classical electrodynamics for given sources. Every radiation formula you will ever use — dipole radiation, antenna theory, Liénard–Wiechert fields of a moving point charge, synchrotron and bremsstrahlung spectra — is a specialization of the retarded potentials derived here. The derivation also settles a foundational question: electromagnetic influences propagate at exactly \( c \), and the fields at \( (\mathbf{r},t) \) are determined entirely by what the sources were doing on the past light cone, never on the future one. Causality is not an add-on assumption; it is a boundary condition selected when we discard the advanced Green's function.

Jefimenko's equations additionally correct a widespread misreading of Faraday's law and the Ampère–Maxwell law. Written as field equations, "changing \( \mathbf{B} \) produces circulating \( \mathbf{E} \)" sounds like causation between fields. Jefimenko's form shows that both \( \mathbf{E} \) and \( \mathbf{B} \) are generated directly by charges and currents (and their time derivatives) at retarded times; the field–field relations are consistency conditions, not causal mechanisms. That conceptual clarity matters when reasoning about electromagnetic induction, radiation reaction, and the near-field/far-field split in real devices.

Assumptions

Microscopic Maxwell equations in vacuum, with prescribed sources \( \rho(\mathbf{r},t) \), \( \mathbf{J}(\mathbf{r},t) \).If sources respond dynamically to the fields they create (plasmas, radiating charges with back-reaction), the equations below remain valid at each instant but no longer constitute a closed solution — they become one half of a self-consistent system. In a linear dispersive medium \( c \) is frequency dependent and the sharp retarded kernel \( \delta(t - t' - R/c) \) is destroyed.

Lorenz gauge, \( \nabla\cdot\mathbf{A} + \frac{1}{c^2}\frac{\partial \varphi}{\partial t} = 0 \) (prior result: potentials and gauge freedom).In another gauge (e.g. Coulomb gauge) the potentials obey different, coupled equations and \( \varphi \) can even respond instantaneously; the fields \( \mathbf{E},\mathbf{B} \) are gauge invariant and unchanged, but the clean wave-equation structure is lost.

Sources are localized (compact support or sufficiently fast falloff) and the fields vanish at spatial infinity.For infinitely extended sources the retarded integrals may diverge and must be regularized (the switched-on infinite wire in Worked Example 2 converges only because causality cuts the integral off at \( R = ct \)).

Retarded ("no incoming radiation") boundary condition: the homogeneous solution added to the particular integral is chosen to be zero in the remote past.The d'Alembertian has a two-parameter family of Green's functions; keeping the advanced kernel \( \delta(t - t' + R/c) \), or any mixture, also solves the PDE. Drop this condition and the solution acquires source-free waves converging from infinity — mathematically legal, physically the statement of your initial conditions. Wheeler–Feynman absorber theory deliberately uses the half-retarded, half-advanced mixture.

Sources are smooth enough to differentiate under the integral sign, and \( \rho, \mathbf{J} \) obey the continuity equation \( \partial_t \rho + \nabla'\cdot\mathbf{J} = 0 \).Continuity is what makes the retarded pair consistent with the Lorenz condition (Step 9); if it fails, the "solution" violates the gauge constraint and hence Maxwell's equations. For point charges (\( \rho \propto \delta^3 \)) the retardation condition must be unfolded with a Jacobian, producing the Liénard–Wiechert factor \( (1 - \hat{\mathbf{R}}\cdot\mathbf{v}/c)^{-1} \) — naive substitution into Jefimenko's integrals is wrong.

Derivation

Notation: \( \mathbf{R} \equiv \mathbf{r} - \mathbf{r}' \), \( R = |\mathbf{R}| \), \( \hat{\mathbf{R}} = \mathbf{R}/R \) (pointing from source point to field point), and square brackets \( [\,f\,] \equiv f(\mathbf{r}', t_r) \) denote evaluation at the retarded time \( t_r = t - R/c \). Overdots denote \( \partial/\partial t \) at fixed \( \mathbf{r}' \).

\[ \Box\, \varphi = -\frac{\rho}{\varepsilon_0}, \qquad \Box\, \mathbf{A} = -\mu_0 \mathbf{J}, \qquad \Box \equiv \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \]

Starting point, imported from the prior results: substituting \( \mathbf{E} = -\nabla\varphi - \partial_t\mathbf{A} \), \( \mathbf{B} = \nabla\times\mathbf{A} \) into Maxwell's equations and imposing the Lorenz condition decouples the potentials into four identical scalar wave equations. A

\[ \Box\, \psi(\mathbf{r},t) = -f(\mathbf{r},t) \]

All four equations share one structure, so solve the generic prototype once: \( \psi \) stands for \( \varphi \) or any Cartesian component of \( \mathbf{A} \), and \( f \) for \( \rho/\varepsilon_0 \) or \( \mu_0 J_i \). Cartesian components are essential — only they obey scalar wave equations (curvilinear components mix under \( \nabla^2 \)). A

\[ \Box\, G(\mathbf{r},t;\mathbf{r}',t') = -\,\delta^3(\mathbf{r}-\mathbf{r}')\,\delta(t-t'), \qquad \psi(\mathbf{r},t) = \int d^3r' \int dt'\; G(\mathbf{r},t;\mathbf{r}',t')\, f(\mathbf{r}',t') \]

Linearity of \( \Box \) permits superposition: if \( G \) is the response to a unit point flash, integrating \( G \) against the source reproduces the general solution, as verified by applying \( \Box \) under the integral. Translation invariance of free space means \( G \) depends only on \( \mathbf{R} \) and \( \tau = t - t' \). B

\[ G(\mathbf{R},\tau) = \int_{-\infty}^{\infty} \frac{d\omega}{2\pi}\, e^{-i\omega \tau}\, G_\omega(\mathbf{R}) \;\;\Longrightarrow\;\; \left( \nabla^2 + k^2 \right) G_\omega = -\,\delta^3(\mathbf{R}), \qquad k = \frac{\omega}{c} \]

Fourier transform in time: \( \partial_t^2 \to -\omega^2 \), converting the wave operator into the Helmholtz operator at each frequency. Legal because \( G \) is a tempered distribution in \( \tau \). B

\[ \frac{1}{R}\frac{d^2}{dR^2}\left( R\, G_\omega \right) + k^2 G_\omega = 0 \;\;(R>0) \;\;\Longrightarrow\;\; G_\omega = \frac{C_+ e^{+ikR} + C_- e^{-ikR}}{4\pi R}, \qquad C_+ + C_- = 1 \]

Spherical symmetry of the point source: with \( u = R\,G_\omega \), the radial Laplacian collapses to \( u'' + k^2 u = 0 \), solved by complex exponentials. The normalization \( C_+ + C_- = 1 \) follows by integrating the Helmholtz equation over a vanishing ball around \( \mathbf{R}=0 \): as \( kR \to 0 \) the solution must match the Coulomb Green's function \( 1/(4\pi R) \) of \( \nabla^2 G = -\delta^3 \). B

\[ C_+ = 1, \;\; C_- = 0: \qquad G_\omega(\mathbf{R}) = \frac{e^{ikR}}{4\pi R} \]

Causal boundary condition. Reassembled with \( e^{-i\omega t} \), the \( e^{+ikR} \) branch is an outgoing spherical wave (phase fronts move toward larger \( R \) as \( t \) increases); \( e^{-ikR} \) is an incoming wave that would make the response precede the flash. Physics, not mathematics, makes this choice — it encodes "no incoming radiation." B

\[ G_{\text{ret}}(\mathbf{R},\tau) = \int_{-\infty}^{\infty} \frac{d\omega}{2\pi}\, \frac{e^{-i\omega\left(\tau - R/c\right)}}{4\pi R} = \frac{\delta\!\left( \tau - \dfrac{R}{c} \right)}{4\pi R} \]

Inverse Fourier transform: the frequency integral of \( e^{-i\omega x}/2\pi \) is \( \delta(x) \) with \( x = \tau - R/c \). The unit flash at \( (\mathbf{r}',t') \) is felt at \( \mathbf{r} \) only at the single instant \( t = t' + R/c \) — a sharp spherical shell expanding at \( c \), with amplitude falling as \( 1/R \). (Sharpness of the shell — Huygens' principle — is special to three spatial dimensions; in 2D the kernel has a wake.) C

\[ \varphi(\mathbf{r},t) = \frac{1}{4\pi\varepsilon_0} \int \frac{[\rho]}{R}\, d^3r', \qquad \mathbf{A}(\mathbf{r},t) = \frac{\mu_0}{4\pi} \int \frac{[\mathbf{J}]}{R}\, d^3r' \]

Insert \( G_{\text{ret}} \) into Step 3 and perform the \( t' \) integral against the delta function, which pins \( t' = t - R/c \equiv t_r \). These are the retarded potentials: Coulomb's and the vector potential's static forms, with each volume element read off at its own retarded time. A

\[ \nabla\cdot\mathbf{A} + \frac{1}{c^2}\frac{\partial \varphi}{\partial t} = \frac{\mu_0}{4\pi} \int \frac{1}{R} \left( \left[ \nabla'\cdot\mathbf{J} \right] + \left[ \dot{\rho} \right] \right) d^3r' = 0 \]

Consistency check (required — Step 1 assumed the Lorenz condition, so the solution must return it). Using \( \nabla t_r = -\hat{\mathbf{R}}/c = -\nabla' t_r \) and integrating by parts to convert \( \nabla \)-derivatives at the field point into \( \nabla' \)-derivatives at the source, the integrand collapses onto the retarded continuity equation \( [\nabla'\cdot\mathbf{J} + \dot\rho] = 0 \). Charge conservation is exactly what keeps the retarded pair inside the Lorenz gauge. C

\[ \nabla [\rho] = [\dot{\rho}]\, \nabla t_r = -\,[\dot{\rho}]\, \frac{\hat{\mathbf{R}}}{c}, \qquad \nabla\!\left( \frac{1}{R} \right) = -\frac{\hat{\mathbf{R}}}{R^2} \;\;\Longrightarrow\;\; \nabla \frac{[\rho]}{R} = -\,\frac{[\rho]\,\hat{\mathbf{R}}}{R^2} - \frac{[\dot{\rho}]\,\hat{\mathbf{R}}}{cR} \]

The key lemma for differentiating retarded integrands: \( \nabla \) at fixed \( t \) acts both on the explicit \( 1/R \) and, through \( t_r(\mathbf{r}) \), on the argument of the source — moving the field point changes which past instant is sampled. Product rule plus chain rule; forgetting the second channel is the classic error. B

\[ \mathbf{E} = -\nabla\varphi - \frac{\partial \mathbf{A}}{\partial t} = \frac{1}{4\pi\varepsilon_0} \int \left[ \frac{[\rho]\,\hat{\mathbf{R}}}{R^2} + \frac{[\dot{\rho}]\,\hat{\mathbf{R}}}{cR} - \frac{[\dot{\mathbf{J}}]}{c^2 R} \right] d^3r' \]

Assemble: Step 10 evaluates \( -\nabla\varphi \); the \( \partial_t \) of \( \mathbf{A} \) passes through the integral to give \( [\dot{\mathbf{J}}] \) (at fixed \( \mathbf{r},\mathbf{r}' \), \( \partial t_r/\partial t = 1 \)), and \( \mu_0 = 1/(\varepsilon_0 c^2) \) unifies the prefactor. A

\[ \nabla \times \frac{[\mathbf{J}]}{R} = \frac{1}{R}\left( \nabla t_r \times [\dot{\mathbf{J}}] \right) - [\mathbf{J}] \times \nabla\!\left(\frac{1}{R}\right) = \frac{[\dot{\mathbf{J}}] \times \hat{\mathbf{R}}}{cR} + \frac{[\mathbf{J}] \times \hat{\mathbf{R}}}{R^2} \]

Curl of the retarded integrand, componentwise: \( \varepsilon_{ijk}\partial_j [J_k] = \varepsilon_{ijk} (\partial_j t_r) [\dot{J}_k] \) since \( \mathbf{J} \) depends on \( \mathbf{r} \) only through \( t_r \); then \( \nabla t_r = -\hat{\mathbf{R}}/c \) and \( -\hat{\mathbf{R}} \times [\dot{\mathbf{J}}] = [\dot{\mathbf{J}}] \times \hat{\mathbf{R}} \). B

\[ \mathbf{B} = \nabla \times \mathbf{A} = \frac{\mu_0}{4\pi} \int \left[ \frac{[\mathbf{J}] \times \hat{\mathbf{R}}}{R^2} + \frac{[\dot{\mathbf{J}}] \times \hat{\mathbf{R}}}{cR} \right] d^3r' \]

Insert Step 12 into the curl of the retarded \( \mathbf{A} \). The first term is the Biot–Savart law with retarded current; the second is its genuinely dynamical correction, surviving at large \( R \) as radiation. A

Result

\[ \varphi(\mathbf{r},t) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\mathbf{r}',t_r)}{R}\, d^3r', \qquad \mathbf{A}(\mathbf{r},t) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}',t_r)}{R}\, d^3r' \] \[ \mathbf{E}(\mathbf{r},t) = \frac{1}{4\pi\varepsilon_0} \int \left[ \frac{\rho(\mathbf{r}',t_r)}{R^2}\,\hat{\mathbf{R}} + \frac{\dot{\rho}(\mathbf{r}',t_r)}{cR}\,\hat{\mathbf{R}} - \frac{\dot{\mathbf{J}}(\mathbf{r}',t_r)}{c^2 R} \right] d^3r' \] \[ \mathbf{B}(\mathbf{r},t) = \frac{\mu_0}{4\pi} \int \left[ \frac{\mathbf{J}(\mathbf{r}',t_r)}{R^2} + \frac{\dot{\mathbf{J}}(\mathbf{r}',t_r)}{cR} \right] \times \hat{\mathbf{R}}\; d^3r', \qquad t_r = t - \frac{R}{c} \]

Reading. The potentials are the static Coulomb and vector-potential integrals with one amendment: each source element speaks from its own past, delayed by the light-travel time \( R/c \). The fields (Jefimenko's equations) say that \( \mathbf{E} \) is built from the retarded charge density (Coulomb-like, \( 1/R^2 \)), the retarded rate of change of charge density (\( 1/R \)), and the retarded acceleration of currents (\( 1/R \)); \( \mathbf{B} \) is retarded Biot–Savart plus a \( \dot{\mathbf{J}} \) radiation term. Only the \( 1/R \) terms carry energy to infinity, since the Poynting flux \( \sim E B R^2 \) then survives as \( R \to \infty \). Everything is causal: no term references the present or future of the source.

Units check. Scalar potential: \( \frac{1}{4\pi\varepsilon_0}\frac{\rho\, d^3r'}{R} \sim \frac{\mathrm{N\,m^2}}{\mathrm{C^2}} \cdot \frac{\mathrm{C\,m^{-3}\cdot m^3}}{\mathrm{m}} = \mathrm{\frac{N\,m}{C}} = \mathrm{V} \). ✓ In \( \mathbf{E} \): first term \( \frac{1}{4\pi\varepsilon_0}\frac{\rho}{R^2} d^3 r' \sim \mathrm{V/m} \); second term gains \( \dot{\rho} \sim \rho\,\mathrm{s^{-1}} \) and loses one \( \mathrm{m^{-1}} \), and dividing by \( c \; (\mathrm{m\,s^{-1}}) \) restores \( \mathrm{V/m} \); third term \( \dot{J}/c^2 R \sim \mathrm{(A\,m^{-2}\,s^{-1})\,m^3 / (m^3\,s^{-2}\,m)} \cdot \frac{1}{4\pi\varepsilon_0} \), and since \( \mathrm{A} = \mathrm{C\,s^{-1}} \) this is again \( \mathrm{V/m} \). ✓ In \( \mathbf{B} \): \( \frac{\mu_0}{4\pi} \frac{J}{R^2} d^3 r' \sim \mathrm{\frac{T\,m}{A}} \cdot \mathrm{\frac{A\,m^{-2}\,m^3}{m^2}} = \mathrm{T} \), and the \( \dot{J}/(cR) \) term matches identically. ✓

Limiting cases

Statics (\( \dot{\rho} = \dot{\mathbf{J}} = 0 \)): retardation is invisible because nothing changes; \( \mathbf{E} \to \) Coulomb's law, \( \mathbf{B} \to \) Biot–Savart law, exactly.
Quasistatics (\( R \ll c\,T \) for characteristic timescale \( T \)): Taylor-expanding in \( R/c \), the first-order retardation corrections to \( \mathbf{E} \) cancel between the \( [\rho] \) and \( [\dot{\rho}] \) terms — the instantaneous Coulomb field is accurate to \( \mathcal{O}\!\left((R/cT)^2\right) \). This is why circuit theory and magnetostatics work far beyond their nominal remit (Problem 5).
Far zone (\( R \gg cT \) and \( R \gg \) source size): the \( 1/R^2 \) terms die; \( \mathbf{E} \to -\frac{1}{4\pi\varepsilon_0 c^2 R}\int [\dot{\mathbf{J}}]_\perp\, d^3r' \) and \( \mathbf{B} \to \hat{\mathbf{R}}\times\mathbf{E}/c \): transverse radiation fields, mutually perpendicular, in ratio \( E/B = c \).
Harmonic sources (\( \rho, \mathbf{J} \propto e^{-i\omega t} \)): retardation becomes the phase factor \( e^{ikR} \) and the potentials reduce to Helmholtz integrals \( \int f(\mathbf{r}') \frac{e^{ikR}}{R} d^3r' \) — the workhorse of antenna theory; \( kR \ll 1 \) and \( kR \gg 1 \) recover the near- and far-zone limits.
Point charge in prescribed motion: careful evaluation of the delta-function sources (with the retardation Jacobian) yields the Liénard–Wiechert potentials and fields.

Breaks when

Dispersive or absorbing media. The derivation used \( \Box \) with a single, frequency-independent \( c \). In matter, \( n(\omega) \) spreads the sharp kernel \( \delta(t-t'-R/c) \) into a smeared response (with precursors governed by the Sommerfeld–Brillouin analysis); "the" retarded time no longer exists, though signal fronts still travel at the vacuum \( c \).
Boundaries and cavities. The Green's function \( \delta(\tau - R/c)/4\pi R \) is that of free space. Conductors, waveguides, or any boundary condition at finite distance demand a different Green's function (image terms, mode sums); blindly using the free-space retarded integrals violates the boundary conditions.
Self-interacting sources. For a charge whose motion responds to its own field, Jefimenko's equations still give the field of a given trajectory but do not close the dynamics; the self-force problem (Abraham–Lorentz–Dirac) requires regularization and lies outside this derivation.
Strong-field / quantum regime. Above the Schwinger scale (\( E \sim 1.3\times 10^{18}\ \mathrm{V/m} \)) vacuum pair creation makes electrodynamics nonlinear; and for single photons the classical field description itself gives way to QED, where the retarded Green's function is replaced by the Feynman propagator.

Failure modes

The frozen-gradient error. Computing \( \nabla \varphi \) by differentiating only the explicit \( 1/R \) and treating \( \rho(\mathbf{r}',t_r) \) as constant. The retarded time depends on \( \mathbf{r} \), so \( \nabla \) also produces \( [\dot{\rho}]\,\nabla t_r = -[\dot{\rho}]\hat{\mathbf{R}}/c \). Omitting it deletes the entire radiation field.
"Retarded Coulomb's law." Writing \( \mathbf{E} = \frac{1}{4\pi\varepsilon_0}\int \frac{[\rho]\hat{\mathbf{R}}}{R^2} d^3r' \) — i.e. Coulomb with \( t \to t_r \) and nothing else. This is not a solution of Maxwell's equations; the \( [\dot\rho] \) and \( [\dot{\mathbf{J}}] \) terms are mandatory. (Curiously, retarded Biot–Savart plus the \( [\dot{\mathbf{J}}] \) term is exactly \( \mathbf{B} \), but the naive \( 1/R^2 \) piece alone is still wrong.)
Retarding the potentials in the wrong gauge. Applying the \( t \to t_r \) recipe to Coulomb-gauge potentials. The retarded forms solve the Lorenz-gauge equations; in Coulomb gauge \( \varphi \) is the instantaneous Coulomb potential and causality hides in \( \mathbf{A} \).
Sign/direction slip in \( \hat{\mathbf{R}} \). Taking \( \hat{\mathbf{R}} \) to point from field point to source. With \( \mathbf{R} = \mathbf{r}-\mathbf{r}' \), \( \hat{\mathbf{R}} \) points source \( \to \) field, and \( \nabla(1/R) = -\hat{\mathbf{R}}/R^2 \), \( \nabla t_r = -\hat{\mathbf{R}}/c \). One flipped sign turns retarded into advanced structure.
Point-charge substitution. Plugging \( \rho = q\,\delta^3(\mathbf{r}'-\mathbf{w}(t_r)) \) into the retarded integrals and integrating as if \( t_r \) were fixed. The delta's argument depends on \( \mathbf{r}' \) through \( t_r(\mathbf{r}') \); the Jacobian gives the Liénard–Wiechert factor \( 1/(1-\hat{\mathbf{R}}\cdot\mathbf{v}/c) \), and missing it fails even for uniform motion.
Expecting \( \mathbf{E} \) to contain \( [\mathbf{J}] \). Students often "symmetrize" Jefimenko's equations by adding a \( [\mathbf{J}]/R^2 \) term to \( \mathbf{E} \). It is not there: current appears in \( \mathbf{E} \) only through \( [\dot{\mathbf{J}}] \) (any \( [\mathbf{J}] \)-dependence is already encoded in \( \rho \) via continuity).

Discussion

The deepest content of this derivation is that causality enters classical electrodynamics as a choice of Green's function, not as a theorem. The d'Alembertian, being time-reversal symmetric, is equally happy with the advanced kernel \( \delta(\tau + R/c)/4\pi R \), which would make fields depend on the sources' future. We discard it because our experimental situation — sources switched on, no conspiratorial waves arriving from infinity — is time-asymmetric, even though the equations are not. This is the same arrow-of-time structure that appears in the retarded response functions of linear-response theory and in the \( i\epsilon \) prescriptions of quantum field theory; the electromagnetic case is simply where most physicists meet it first.

Jefimenko's equations sharpen the physical reading of Maxwell's equations. In differential form, Faraday's law couples \( \nabla\times\mathbf{E} \) to \( \partial_t\mathbf{B} \) at the same spacetime point — simultaneity that cannot be causation. In Jefimenko's form the only causes are \( [\rho], [\dot\rho], [\mathbf{J}], [\dot{\mathbf{J}}] \) on the past light cone; \( \mathbf{E} \) and \( \mathbf{B} \) are cousins with a common ancestor, not parents of one another. This resolves apparent paradoxes about "which field came first" in induction problems, while leaving all calculations untouched — the field–field form and the source–field form are mathematically equivalent for prescribed sources.

The three-term structure of \( \mathbf{E} \) encodes the near/intermediate/far zone hierarchy quantitatively. For a source of size \( d \), timescale \( T \), observed at distance \( R \): the \( 1/R^2 \) term dominates for \( R \ll cT \) (quasistatic zone), the \( \dot{\rho}, \dot{\mathbf{J}} \) terms take over for \( R \gg cT \), and the ratio of dynamic to static terms is \( \sim R/cT = kR/2\pi \) for harmonic sources — Worked Example 1 makes this concrete. Energy bookkeeping follows: only \( 1/R \) fields give \( |\mathbf{S}| R^2 \not\to 0 \), so radiation is precisely the part of the field sourced by time derivatives. A charge in uniform motion has \( \dot{\mathbf{J}} \neq 0 \) at fixed points of space, yet does not radiate — the \( 1/R \) contributions assemble into a field that merely convects with the charge, a cancellation worth verifying once in a lifetime via Liénard–Wiechert.

Relativistically, the retarded Green's function has the covariant form \( G_{\text{ret}}(x-x') = \frac{1}{2\pi}\,\theta(t-t')\,\delta\!\left( (x-x')^2 \right) \) with \( (x-x')^2 = c^2(t-t')^2 - |\mathbf{r}-\mathbf{r}'|^2 \): support exactly on the past light cone, with the step function \( \theta \) supplying the only frame-dependent-looking ingredient — yet the retarded/advanced split is Lorentz invariant because the light cone's interior sheets cannot be exchanged by orthochronous transformations. In Lorenz gauge the whole derivation compresses to \( \Box A^\mu = -\mu_0 J^\mu \Rightarrow A^\mu(x) = \frac{\mu_0}{4\pi}\int d^4x'\, \frac{\theta(t-t')\,\delta((x-x')^2)}{2\pi}\cdot 2c\, J^\mu(x') \), manifestly covariant since \( \delta((x-x')^2) \) is a scalar and \( J^\mu \) a four-vector. The residual gauge freedom \( A^\mu \to A^\mu + \partial^\mu \chi \) with \( \Box\chi = 0 \) is fixed by the same no-incoming-wave condition that selected retardation. Common misconceptions: (i) retardation does not mean the field of a uniformly moving charge points to where the charge was — the velocity-field terms conspire so \( \mathbf{E} \) points at the present (extrapolated) position; (ii) the Coulomb-gauge scalar potential's instantaneous action does not violate relativity, because \( \varphi \) alone is not observable and the gauge-invariant fields from Jefimenko's equations are strictly retarded; (iii) \( t_r \) is not a single time — every source element has its own, which is why "the field now equals the source pattern a moment ago" fails for extended sources.

Worked examples

Example 1 — When does retardation actually matter? A small sphere at the origin carries charge \( q(t) = q_0 e^{-t/\tau} \) (discharged through a thin radial wire; the wire's contribution to \( \mathbf{E} \) at our field point on the perpendicular axis is negligible for this estimate). Take \( q_0 = 1.0\ \mathrm{nC} \), \( \tau = 10\ \mathrm{ns} \), and evaluate the two scalar-charge terms of Jefimenko's \( \mathbf{E} \) at \( R = 3.0\ \mathrm{m} \), at the moment when \( q(t_r) = q_0 \).

\[ E_1 = \frac{1}{4\pi\varepsilon_0}\,\frac{q(t_r)}{R^2}, \qquad E_2 = \frac{1}{4\pi\varepsilon_0}\,\frac{|\dot{q}(t_r)|}{cR}, \qquad \frac{E_2}{E_1} = \frac{|\dot{q}|\,R}{q\,c} = \frac{R}{c\tau} \]

For a compact source the volume integrals collapse: \( \int [\rho]\, d^3r' = q(t_r) \), \( \int [\dot{\rho}]\, d^3r' = \dot{q}(t_r) \), with \( |\dot q| = q/\tau \) for the exponential. Symbols first: the dimensionless ratio \( R/c\tau \) controls everything. A

\[ E_1 = \left( 8.99\times 10^{9}\ \mathrm{\tfrac{N\,m^2}{C^2}} \right) \frac{1.0\times 10^{-9}\ \mathrm{C}}{(3.0\ \mathrm{m})^2} = 1.0\ \mathrm{V/m} \]

Insert numbers into the static-form term. A

\[ E_2 = \left( 8.99\times 10^{9} \right) \frac{(1.0\times 10^{-9}/10\times 10^{-9})\ \mathrm{A}}{(3.00\times 10^{8}\ \mathrm{m/s})(3.0\ \mathrm{m})} = \frac{8.99\times 10^{9} \times 0.10}{9.0\times 10^{8}} = 1.0\ \mathrm{V/m} \]

The retardation term: \( |\dot q| = q_0/\tau = 0.10\ \mathrm{A} \). It is equal to the Coulomb term because \( c\tau = (3\times10^8)(10^{-8}) = 3.0\ \mathrm{m} = R \): we are sitting exactly on the near/far boundary. A

\[ E_1 \approx E_2 \approx 1.0\ \mathrm{V/m}, \qquad \frac{E_2}{E_1} = \frac{R}{c\tau} = 1.0 \]

Reading. At \( R = c\tau \) the "correction" is as large as the Coulomb term itself — quasistatic intuition has completely expired. At \( R = 30\ \mathrm{cm} \) the ratio would be \( 0.1 \) (quasistatics decent); at \( 30\ \mathrm{m} \), \( 10 \) (radiation zone, Coulomb term negligible).

Units check. \( \mathrm{\frac{N\,m^2}{C^2}}\cdot\mathrm{\frac{A}{(m/s)\,m}} = \mathrm{\frac{N\,m^2}{C^2}}\cdot\mathrm{\frac{C}{m^2}} = \mathrm{\frac{N}{C}} = \mathrm{V/m} \). ✓

Example 2 — Switching on an infinite wire. A neutral infinite straight wire on the \( z \)-axis carries \( I(t) = 0 \) for \( t < 0 \) and \( I(t) = I_0 \) for \( t \geq 0 \). Find \( \mathbf{B} \) and \( \mathbf{E} \) at cylindrical radius \( s \), and evaluate for \( I_0 = 10\ \mathrm{A} \), \( s = 0.30\ \mathrm{m} \), \( t = 2.0\ \mathrm{ns} \).

\[ \mathbf{A}(s,t) = \frac{\mu_0}{4\pi}\,\hat{\mathbf{z}} \int_{-\infty}^{\infty} \frac{I(t_r)}{\sqrt{s^2+z^2}}\, dz, \qquad t_r = t - \frac{\sqrt{s^2+z^2}}{c} \]

Retarded vector potential (Step 8 of the derivation) for a line current; \( \varphi = 0 \) since \( \rho = 0 \). The integrand vanishes unless \( t_r > 0 \), i.e. unless news of the switch-on has arrived from element \( z \): \( \sqrt{s^2+z^2} < ct \). Causality regulates an otherwise divergent integral. B

\[ \mathbf{A}(s,t) = \frac{\mu_0 I_0}{4\pi}\,\hat{\mathbf{z}}\; 2\!\int_{0}^{\sqrt{(ct)^2 - s^2}} \frac{dz}{\sqrt{s^2+z^2}} = \frac{\mu_0 I_0}{2\pi} \ln\!\left( \frac{ct + \sqrt{(ct)^2 - s^2}}{s} \right) \hat{\mathbf{z}} \quad (ct > s) \]

Only \( |z| < \sqrt{(ct)^2 - s^2} \) contributes; the integral is the standard \( \ln(z + \sqrt{s^2+z^2}) \) antiderivative. For \( ct < s \), \( \mathbf{A} = 0 \): no field before light arrives. B

\[ \mathbf{B} = \nabla\times\mathbf{A} = -\frac{\partial A_z}{\partial s}\,\hat{\boldsymbol{\varphi}} = \frac{\mu_0 I_0}{2\pi s}\, \frac{ct}{\sqrt{(ct)^2 - s^2}}\; \hat{\boldsymbol{\varphi}}, \qquad \mathbf{E} = -\frac{\partial \mathbf{A}}{\partial t} = -\frac{\mu_0 I_0 c}{2\pi \sqrt{(ct)^2 - s^2}}\; \hat{\mathbf{z}} \]

Differentiate the closed form: the \( s \)-derivative uses \( \partial_s \sqrt{(ct)^2-s^2} = -s/\sqrt{(ct)^2-s^2} \) and simplifies via \( s^2 + \left[(ct)^2 - s^2\right] = (ct)^2 \). As \( t \to \infty \): \( \mathbf{B} \to \frac{\mu_0 I_0}{2\pi s}\hat{\boldsymbol{\varphi}} \) (magnetostatics recovered) and \( \mathbf{E} \to 0 \). B

\[ ct = (3.00\times 10^{8})(2.0\times 10^{-9}) = 0.60\ \mathrm{m}, \qquad \sqrt{(ct)^2 - s^2} = \sqrt{0.36 - 0.09} = 0.5196\ \mathrm{m} \]

Numbers last. The news arrived at \( s = 0.30\ \mathrm{m} \) at \( t = s/c = 1.0\ \mathrm{ns} \), so at \( 2.0\ \mathrm{ns} \) the field is one nanosecond old. A

\[ B = \frac{(4\pi\times 10^{-7})(10)}{2\pi (0.30)}\cdot\frac{0.60}{0.5196} = (6.67\ \mu\mathrm{T})(1.155) = 7.7\ \mu\mathrm{T}, \quad E = \frac{(2\times 10^{-7})(10)(3.00\times 10^{8})}{0.5196} = 1.15\ \mathrm{kV/m} \]

Evaluate: \( \mu_0/2\pi = 2\times 10^{-7}\ \mathrm{T\,m/A} \). The magnetic field is already within \( 16\% \) of its final static value; the transient \( E_z \), antiparallel to the current, is enormous but collapsing like \( 1/t \) — it is the field that fights the establishment of the current (the flip side of inductive back-EMF). A

\[ B(0.30\ \mathrm{m},\, 2.0\ \mathrm{ns}) = 7.7\ \mu\mathrm{T}\;\hat{\boldsymbol{\varphi}}, \qquad E = -1.15\ \mathrm{kV/m}\;\hat{\mathbf{z}} \]

Reading. Before \( t = s/c \) there is literally nothing; at the light front both fields diverge (an artifact of the instantaneous switch-on — any finite rise time smooths it); afterwards \( \mathbf{B} \) relaxes down to Biot–Savart and \( \mathbf{E} \) decays to zero. Retardation is not a small correction here — it is the entire structure of the answer.

Units check. \( \frac{\mu_0 I_0 c}{2\pi\, \mathrm{m}} \sim \mathrm{\frac{T\,m}{A}\cdot A \cdot \frac{m/s}{m}} = \mathrm{T\,m/s} = \mathrm{V/m} \) (since \( \mathrm{T} = \mathrm{V\,s/m^2} \)). ✓

Problems

A compact source at the origin emits a pulse at \( t = 0 \). An observer sits at \( \mathbf{r} = (90, 120, 0)\ \mathrm{m} \). At what time does the observer's field first respond, and what is the retarded time the observer's field "reads" at the observer's clock time \( t = 1.00\ \mu\mathrm{s} \)?
Solution
Distance: \( R = \sqrt{90^2 + 120^2} = \sqrt{8100 + 14400} = \sqrt{22500} = 150\ \mathrm{m} \). First response at \( t = R/c = 150 / (3.00\times 10^{8}) = 5.00\times 10^{-7}\ \mathrm{s} = 0.500\ \mu\mathrm{s} \). At \( t = 1.00\ \mu\mathrm{s} \), the field depends on the source at \( t_r = t - R/c = 1.00 - 0.50 = 0.50\ \mu\mathrm{s} \). The observer always sees the source half a microsecond in its past.
Verify the static limit numerically. (a) A point charge \( q = 2.0\ \mu\mathrm{C} \) sits at rest; find \( E \) at \( R = 0.50\ \mathrm{m} \) from Jefimenko's equation. (b) A long straight wire carries steady \( I = 2.0\ \mathrm{A} \); find \( B \) at \( s = 5.0\ \mathrm{cm} \).
Solution
(a) With \( \dot\rho = 0 \) and \( \dot{\mathbf{J}} = 0 \), only the first term survives and \( t_r \) is irrelevant (the source never changes): \( E = \frac{1}{4\pi\varepsilon_0}\frac{q}{R^2} = (8.99\times 10^{9})\frac{2.0\times 10^{-6}}{(0.50)^2} = 7.2\times 10^{4}\ \mathrm{V/m} = 72\ \mathrm{kV/m} \), radially outward. (b) Only the \( [\mathbf{J}]\times\hat{\mathbf{R}}/R^2 \) term survives, which is Biot–Savart; for an infinite wire this integrates to \( B = \frac{\mu_0 I}{2\pi s} = \frac{(2\times 10^{-7})(2.0)}{0.050} = 8.0\times 10^{-6}\ \mathrm{T} = 8.0\ \mu\mathrm{T} \), azimuthal. Jefimenko contains all of electro- and magnetostatics as the frozen-source special case.
A small antenna's charge distribution oscillates harmonically at \( f = 100\ \mathrm{MHz} \). Estimate the ratio of the \( [\dot{\rho}]/(cR) \) term to the \( [\rho]/R^2 \) term in \( \mathbf{E} \) at \( R = 10\ \mathrm{m} \), and classify the zone. At what distance are the two terms equal?
Solution
For \( \rho \propto e^{-i\omega t} \), \( |\dot\rho| = \omega|\rho| \), so the ratio is \( \frac{|\dot\rho|/(cR)}{|\rho|/R^2} = \frac{\omega R}{c} = kR \). Numerically \( \omega = 2\pi f = 6.28\times 10^{8}\ \mathrm{s^{-1}} \), so \( kR = \frac{(6.28\times 10^{8})(10)}{3.00\times 10^{8}} = 20.9 \). The dynamic term dominates by a factor \( \sim 21 \): far (radiation) zone. Equality at \( kR = 1 \): \( R = c/\omega = \lambda/2\pi = \frac{3.00}{6.28} = 0.48\ \mathrm{m} \) — the "reduced wavelength" marks the near/far boundary.
For the switched-on wire of Worked Example 2 with \( I_0 = 5.0\ \mathrm{A} \), an observer at \( s = 0.90\ \mathrm{m} \): (a) when does \( \mathbf{B} \) first become nonzero? (b) Compute \( B \) at \( t = 5.0\ \mathrm{ns} \) and compare with the static value.
Solution
(a) At \( t = s/c = 0.90/(3.00\times 10^{8}) = 3.0\ \mathrm{ns} \). Strictly zero before — a direct display of causality. (b) \( ct = (3.00\times 10^{8})(5.0\times 10^{-9}) = 1.50\ \mathrm{m} \); \( \sqrt{(ct)^2 - s^2} = \sqrt{2.25 - 0.81} = \sqrt{1.44} = 1.20\ \mathrm{m} \). Static value: \( B_\infty = \frac{\mu_0 I_0}{2\pi s} = \frac{(2\times 10^{-7})(5.0)}{0.90} = 1.11\ \mu\mathrm{T} \). Then \( B = B_\infty \cdot \frac{ct}{\sqrt{(ct)^2 - s^2}} = 1.11 \times \frac{1.50}{1.20} = 1.11 \times 1.25 = 1.4\ \mu\mathrm{T} \). Note the field approaches its static value from above: 2 ns after first arrival it is still \( 25\% \) larger than \( B_\infty \) (it diverged at the light front) and decays monotonically toward the Biot–Savart value as \( t \to \infty \).
(Quasistatic expansion.) Expand the charge terms of Jefimenko's \( \mathbf{E} \) to second order in the retardation \( R/c \) and show the first-order terms cancel, so \( \mathbf{E} \approx \mathbf{E}_{\text{Coulomb}}(t) + \mathcal{O}(c^{-2}) \). Then estimate the fractional error of the instantaneous Coulomb field for a source with characteristic timescale \( T = 1.0\ \mu\mathrm{s} \) observed at \( R = 0.10\ \mathrm{m} \).
Solution
Expand about the present: \( \rho(t_r) = \rho(t) - \frac{R}{c}\dot\rho(t) + \frac{R^2}{2c^2}\ddot\rho(t) - \dots \) and \( \dot\rho(t_r) = \dot\rho(t) - \frac{R}{c}\ddot\rho(t) + \dots \). Insert: \[ \frac{\rho(t_r)}{R^2} + \frac{\dot\rho(t_r)}{cR} = \frac{\rho(t)}{R^2} - \frac{\dot\rho(t)}{cR} + \frac{\ddot\rho(t)}{2c^2} + \frac{\dot\rho(t)}{cR} - \frac{\ddot\rho(t)}{c^2} + \dots = \frac{\rho(t)}{R^2} - \frac{\ddot\rho(t)}{2c^2} + \dots \] The \( \dot\rho \) terms cancel identically: the first-order retardation correction to the Coulomb field vanishes, and the leading error is second order, \( \sim \ddot\rho/c^2 \sim \rho/(cT)^2 \) relative to \( \rho/R^2 \), i.e. fractional error \( \sim (R/cT)^2 \). Numerically: \( \frac{R}{cT} = \frac{0.10}{(3.00\times 10^{8})(1.0\times 10^{-6})} = \frac{0.10}{300} = 3.3\times 10^{-4} \), so the error is \( \sim (3.3\times 10^{-4})^2 \approx 1.1\times 10^{-7} \) — one part in ten million. This accidental first-order cancellation is why "instantaneous" circuit and electrostatic reasoning is superbly accurate at bench scales and MHz timescales, and why retardation went unnoticed until Hertz.